Can You State The Possible Number Of Imaginary Zeros Of F(x) = 10x3 - 4x2 + 2x - 6?

Given a cubic equation of the form ax³+bx²+cx+d=0, an annoyingly complicated formula for the "discriminant" will tell you how many real roots the equation has. The discriminant is 18abcd-4(ac³+b³d)+b²c²-27a²d². If it is <0, there are two complex roots and one real root. If it is =0, all roots are real and at least two of them are equal. If the discriminant is >0, there are three unequal real roots.

Another quick, but not so comprehensive, test is to look at b²-3ac. If this value is <0, the equation has only one real root. If this value is =0 or >0, something can be said about the number of inflection points in the graph of ax³+bx²+cx+d, but that is all.

So, in your F(x) = 10x³ - 4x² + 2x - 6, the value of b²-3ac is (-4)²-3(10)(2)=16-60=-44. Thus, the function F(x) is monotonic (has no inflection points) and there is only one real root. The other two roots must be complex, i.e. they have a non-zero imaginary part.

You know the calculator gives you the answers if you plug it in to a calculator